help in trigo..........
Attachments:
Raja395:
1 + tan²A = sec²A
1 + cot²A = cosec²A
Answers
Answered by
0
You must be Knowing the following formulas:
★ (1 + tan²A) = sec²A
★ (1 + cot²A) = cosec²A
So, just apply it in the given question.
We get,
→ (tanA/sec²A) + (cotA/cosec²A)
(Now change all "tanA", "cotA", "secA", & "cosecA" in terms of "SinA" & "CosA".
It would be easier.)
→ (sinA/cosA)(cos²A) + (cosA/sinA)(sin²A)
= sinA cosA + cosA sinA
= 2 sinA cosA
= sin 2A
Thankyou!!!
★ (1 + tan²A) = sec²A
★ (1 + cot²A) = cosec²A
So, just apply it in the given question.
We get,
→ (tanA/sec²A) + (cotA/cosec²A)
(Now change all "tanA", "cotA", "secA", & "cosecA" in terms of "SinA" & "CosA".
It would be easier.)
→ (sinA/cosA)(cos²A) + (cosA/sinA)(sin²A)
= sinA cosA + cosA sinA
= 2 sinA cosA
= sin 2A
Thankyou!!!
Answered by
0
Answer:
[tanA/(1+tan^2A)^2]+[cotA/(1+cot^2A)^2] = sinAcosA
1+tan^2A=sec^2A ..(1)
1+cot^2A=cosec^2A ..(2)
now, put these value in question
(tanA/sec^4A)+(cotA/cosec^4A)
convert into sin and cos ratios
=sinAcos^3A+cosAsin^3A
=(sinAcosA)(sin^2A=cos^2A)
∴sin^2A+cos^2A=1
=sinAcosA
Similar questions